Determine if the points (1,5), (2,3) and (– 2, – 11) are collinear.

Solution:

The given points are :

A(1, 5), B(2, 3) and C(–2, –11).

Let us calculate the distance : AB, BC and CA by using distance formula.

$A B=\sqrt{(2-1)^{2}+(3-5)^{2}}=\sqrt{(1)^{2}+(-2)^{2}}$

$=\sqrt{1+4}=\sqrt{5}$ units

$B C=\sqrt{(-2-2)^{2}+(-11-3)^{2}}=\sqrt{(-4)^{2}+(-14)^{2}}$

$=\sqrt{16+196}=\sqrt{212}=2 \sqrt{53}$ units

$\mathrm{CA}=\sqrt{(-2-1)^{2}+(-11-5)^{2}}$

$=\sqrt{(-3)^{2}+(-16)^{2}}=\sqrt{9+256}=\sqrt{265}$

$=\sqrt{5} \times \sqrt{53}$ units

From the above we see that : $\mathrm{AB}+\mathrm{BC} \neq \mathrm{CA}$

Hence the above stated points $A(1,5), B(2,3)$ and $C(-2,-11)$ are not collinear.